OFFSET
1,1
COMMENTS
First prime p such that the expansion of 1/p has period (p-1)/(2*n) in base 16. Also the first prime p such that {k/p : 1 <= k <= p-1} has 2*n different cycles when written out in base 16.
Since ord(a^m,k) = ord(a,k)/gcd(m,ord(a,k)) for gcd(a,k) = 1, we have that (p-1)/ord(16,p) = ((p-1)/ord(2,p)) * gcd(4,ord(2,p)) is always even. Here ord(a,k) is the multiplicative order of a modulo k.
LINKS
Jean-François Alcover, Table of n, a(n) for n = 1..1000
EXAMPLE
In the following examples let () denote the reptend. The prime numbers themselves and the fractions are written out in decimal.
The base-16 expansion of 1/3 is 0.(5), so the reptend has length 1 = (3-1)/2. Also, the base-16 expansions of 1/3 = 0.(5) and 2/3 = 0.(A) have two cycles 5 and A. 3 is the smallest such prime, so a(1) = 3.
The base-16 expansion of 1/5 is 0.(3), so the reptend has length 1 = (5-1)/4. Also, the base-16 expansions of 1/5 = 0.(3), 2/5 = (0.6), 3/5 = 0.(9) and 4/5 = 0.(C) have four cycles 3, 6, 9 and A. 5 is the smallest such prime, so a(2) = 5.
The base-16 expansion of 1/31 is 0.(08421), so the reptend has length 5 = (31-1)/6. Also, the base-16 expansions of 1/31, 2/31, ..., 30/31 have six cycles 08421, 18C63, 294A5, 39CE7, 5AD6B and 7BDEF. 31 is the smallest such prime, so a(3) = 31.
MATHEMATICA
a[n_] := a[n] = For[p = 2, True, p = NextPrime[p], If[MultiplicativeOrder[16, p] == (p-1)/(2n), Return[p]]];
Table[Print[n, " ", a[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Nov 24 2024 *)
PROG
(PARI) a(n, {base=16}) = forprime(p=2, oo, if((base%p) && znorder(Mod(base, p)) == (p-1)/(n * if(issquare(base), 2, 1)), return(p)))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, May 13 2024
STATUS
approved